Answer
$$y=2x+1$$
Work Step by Step
$a)$ To find the slope $m$ of the tangent line, use the following formula:
$$m = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$ individually, then take the quotient to obtain $m$.
$$\frac{dy}{dt} = \frac{d}{dt}(t^2+2) = 2t$$$$\frac{dx}{dt} = \frac{d}{dt}(1+\ln t) = \frac{1}{t}$$$$m = \frac{2t}{(\frac{1}{t})}=2t^2$$
Since we want the slope at the point $(1,3),$ find the value of $t$ at this point. The $x$-coordinate is $1,$ so set the equation for $x$ equal to $1$ and solve for $t$.
$$x = 1 + \ln t = 1$$$$\ln t = 0$$$$t = 1$$
Evaluate $m$ at $t=1$.
$$m=2(1)^2=2$$
Now use the point-slope formula to find the equation of the tangent line at $(1,3)$:
$$y-y_{0}=m(x-x_{0})$$$$y-3=2(x-1)$$$$y=2x+1$$
$b)$ To eliminate the parameter, first solve for $t$ in terms of $x$:
$$x = 1 + \ln t$$$$t = e^{x-1}$$
Now plug this expression for $t$ into the equation for $y$:
$$y=t^2+2$$$$y =e^{2x-2}+2$$
Take the derivative with respect to $x$ to find the slope:
$$\frac{dy}{dx}=2e^{2x-2}$$
Evaluate $\frac{dy}{dx}$ at the point $(1,3)$ to find $m$ at that point:
$$m=2e^{2(1)-2}=2$$
Again, use the point-slope formula to find the equation of the tangent line at $(1,3)$ to get:
$$y=2x+1$$
